KC Sinha Mathematics Solution Class 12 Chapter 11 अवकलन (Differentiation) Exercise 11.3 (Q7-Q10)

Exercise 11.3

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Question 7

यदि (If) $y=x+\frac{1}{x}_{+\frac{1}{x}_{+\frac{1}{x}_+\dots \text{ to } \infty}}$ सिद्ध करें कि( prove that) $\left(x^{2}-y^{2}+3\right) \frac{d y}{d x}=1$
Sol :
$y=x+\frac{1}{x}_{+\frac{1}{x}_{+\frac{1}{x}_+\dots \text{ to } \infty}}$

$y=x+\frac{1}{y}$

Differentiating with respect to x

$\frac{d y}{d x}=1-\frac{1}{y^{2}} \cdot \frac{d y}{d x}$

$\frac{d y}{dx}+\frac{1}{y^{2}} \frac{d y}{d x}=1$

$\left[1+\frac{1}{y^{2}}\right] \frac{d y}{d x}=1$

$\left[y^{2}-2 \cdot y \cdot \frac{1}{y}+\frac{1}{y^{2}}-y^{2}+3\right] \frac{d y}{d x}=1$

$\left[\left(y-\frac{1}{y}\right)^{2}-y^{2}+3\right] \frac{d y}{d x}=1$

$\left[x^{2}-y^{2}+3\right] \frac{d y}{d x}=1$

Question 8

यदि (If) $y=\sqrt{x+\sqrt{x+\sqrt{x+\ldots . \text { to } \infty}}}$ सिद्ध करे कि (prove that) $\frac{d y}{d x}=\frac{1}{2 y-1}$
Sol :
$y=\sqrt{x+\sqrt{x+\sqrt{x+\ldots . \text { to } \infty}}}$

$y=\sqrt{x+y}$

Squaring both sides

$y^{2}=x+y$

Differentiating with respect to x

$2 y.\frac{d y}{dx}=1+\frac{d y}{d x}$

$2 y \frac{d y}{d x}-\frac{d y}{d x}=1$

$(2 y-1) \frac{d y}{d x}=1$

$\frac{d y}{d x}=\frac{1}{2 y-1}$

Question 9

यदि (If) $\sqrt{1-x^{2}}+\sqrt{1-y^{2}}=a(x-y)$ सिद्ध करें कि (prove that) $\frac{d y}{d x}=\sqrt{\frac{1-y^{2}}{1-x^{2}}}$
Sol :
$\sqrt{1-x^{2}}+\sqrt{1-y^{2}}=a(x-y)$

Differentiating with respect to x

$\frac{1}{2 \sqrt{1-x^{2}}} \times(-2 x)+\frac{1}{2 \sqrt{1-y^{2}}} \times(-2 y) \cdot \frac{d y}{d x}=a\left[1-\frac{dy}{d x}\right]$

$\frac{-x}{\sqrt{1-x^{2}}}-\frac{y}{\sqrt{1-y^{2}}} \frac{d y}{d x}=a-\frac{a d y}{d x}$

$a \frac{d y}{d x}-\frac{y}{\sqrt{1-y^{2}}} \frac{d y}{d x}=a+\frac{x}{\sqrt{1-x^{2}}}$

$\left[a-\frac{y}{\sqrt{1-y^{2}}}\right] \frac{d y}{d x}=a+\frac{x}{\sqrt{1-x^{2}}}$

$\left[\frac{\sqrt{1-x^{2}}+\sqrt{1-y^{2}}}{x-y}-\frac{y}{\sqrt{1-y^{2}}}\right] \frac{d y}{d x}=\frac{\sqrt{1-x^{2}}+\sqrt{1-y^{2}}}{x-y}+\frac{x}{\sqrt{1-x^{2}}}$

$\left[\frac{\sqrt{1-x^{2}} \sqrt{1-y^{2}}+1-y^{2}-x y+y^{2}}{(x-y) \sqrt{1-y^{2}}}\right] \frac{d y}{d x}=\frac{1-x^{2}+\sqrt{1-x^{2}} \sqrt{1-y^{2}}+x^{2}-xy}{(x-y) \sqrt{1-x^{2}}}$

$\left[\frac{\sqrt{1-x^{2}} \sqrt{1-y^{2}}-x y+1}{\sqrt{1-y^{2}}}\right] \frac{d y}{d x}=\frac{\sqrt{1-x^{2}} \sqrt{1-y^{2}}-x y+1}{\sqrt{1-x^{2}}}$

$\frac{dy}{d x}=\frac{\sqrt{1-y^{2}}}{\sqrt{1-x^{2}}}=\sqrt{\frac{1-y^{2}}{1-x^{2}}}$

Question 10

यदि (If) $y=x+\frac{1}{x}$ सिद्ध करें कि (prove that) $x \frac{d y}{d x}+y=2 x$ $x \frac{d y}{d x}+y=2 x$
Sol :
$y=x+\frac{1}{x}$

$y=\frac{x^{2}+1}{x}$

$x y=x^{2}+1$

Differentiating with respect to x

$1 \cdot y+x \frac{d y}{d x}=2 x+0$

$x \frac{d y}{d x}+y=2 x$